Here is my question
A particle attached to a string of length $2$ m is given an initial velocity of $6 \text {m/s}$. The string is attached to a peg and, as the particle rotates about the peg, the string winds around the peg. What length of string has wound around the peg when the velocity of the particle is $20 \text {m/s}$?
I'm training for physics olympiads. That's why I need help.
Good question. The intuitive approach is to apply conservation of momentum, which would give you a length of $3/5$ meters. But why is angular momentum conserved in this system? After all, linear momentum is not conserved, and there is no obvious rotational symmetry to the system.
There may be a clever way of arguing that angular momentum must be conserved, but it's also straightforward to derive this fact. We can parameterized the system by a one-dimensional configuration space representing the angle $\theta$ by which the string has wound around the peg. As the string winds, it shortens: let $l(\theta)$ be the (monotonic decreasing, with $l(0) = 2$, but otherwise unknown) length of the string after it has wound by $\theta$.
Assuming the particle has mass $m$, the kinetic energy of the particle is $$T = \frac{1}{2}m [2\pi l(\theta) \dot \theta]^2;$$ since there is no potential energy the Euler-Lagrange equations are $$\frac{d}{dt} \left(4\pi^2 ml(\theta)^2\dot\theta\right) - 4\pi^2 m l(\theta)\dot\theta^2 l'(\theta)=0$$ or \begin{align*} 8\pi^2 m l(\theta)l'(\theta)\dot\theta^2 + 4\pi^2ml(\theta)^2\ddot\theta - 4\pi^2 ml(\theta)\dot\theta^2l'(\theta)&=0\\ l'(\theta)\dot\theta^2 + l(\theta)\ddot\theta &=0\\ \frac{d}{dt}\left(l(\theta)\dot\theta\right)&=0, \end{align*} and we see that $3/5$ meters is indeed the correct approach.